Question

Which of the following second degree equation represents a pair of straight lines
A. \[{{x}^{2}}-xy-{{y}^{2}}=1\]
B. \[-{{x}^{2}}+xy-{{y}^{2}}=1\]
C. \[4{{x}^{2}}-4xy+{{y}^{2}}=4\]
D. \[{{x}^{2}}+{{y}^{2}}=4\]

Answer 1

Hint: The condition for straight line is that $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$.Comparing the given equation with the general equation we can easily determine the value of the coefficients.

Formula Used:$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$
$a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$



Complete step by step solution:The given equation is \[{{x}^{2}}-xy-{{y}^{2}}=1\]
The general equation of straight line is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$.
Comparing the given equation with the equation we can get the following values.
\[\text{a}=1,h=-\dfrac{1}{2},b=-1,c=-1,f=0,g=0\]
The condition for straight line is given as follows-
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$
Putting the values of $a,b,c,g,f$ in the condition for straight line we get-
$ abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}} \\ $
 $ 1\times \left( -1 \right)\times \left( -1 \right)+2\times 0\times \left( 0 \right)\times {}^{-1}/{}_{2}-1{{\left( 0 \right)}^{2}}-\left( -1 \right)\times {{\left( 0 \right)}^{2}}-\left( -1 \right){{(-\dfrac{1}{2})}^{2}} \\$
 $ +1+0+0+0-\dfrac{1}{4} \\ $
 $ \dfrac{3}{4}\ne 0 \\ $

Thus this equation does not represent a pair of straight lines.
The given equation is \[-{{x}^{2}}+xy-{{y}^{2}}=1\]
The general equation of straight line is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$.
Comparing the given equation with the equation we can get the following values.
\[\text{a}=-1,h=\dfrac{1}{2},b=-1,c=-1,f=0,g=0\]
The condition for straight line is given as follows-
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$
Putting the values of $a,b,c,g,f$ in the condition for straight line we get-

  $ abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}} \\ $
 $ (-1)\times \left( -1 \right)\times \left( -1 \right)+2\times 0\times \left( 0 \right)\times {}^{1}/{}_{2}-(-1){{\left( 0 \right)}^{2}}-\left( -1 \right)\times {{\left( 0 \right)}^{2}}-\left( -1 \right){{(\dfrac{1}{2})}^{2}} \\ $
 $ -1+0+0+0+\dfrac{1}{4} \\ $
 $\dfrac{-3}{4}\ne 0 \\ $

Thus this equation does not represent a pair of straight lines.
The given equation is \[4{{x}^{2}}-4xy+{{y}^{2}}=4\]
The general equation of straight line is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$.
Comparing the given equation with the equation we can get the following values.
\[\text{a}=4,h=-2,b=1,c=-4,f=0,g=0\]
The condition for straight line is given as follows-
$abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$
Putting the values of $a,b,c,g,f$ in the condition for straight line we get-

  $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}} \\ $
 $ (4)\times \left( 1 \right)\times \left( -4 \right)+2\times 0\times \left( 0 \right)\times (-2)-(4){{\left( 0 \right)}^{2}}-\left( 1 \right)\times {{\left( 0 \right)}^{2}}-\left( -4 \right){{(-2)}^{2}} \\ $
 $-16+0+0+0+16 \\ $
 $ =0 \\$

Thus this equation represents a pair of straight lines.



Option ‘C’ is correct



Note: The another method to solve this problem is the matrix method. The condition for straight line is that the determinant of the matrix is zero.